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anonymous

  • 5 years ago

Is anyone good with vector space?

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  1. nowhereman
    • 5 years ago
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    They are my homies :-D

  2. anonymous
    • 5 years ago
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    ha, ha, ha. So are u down with them as well?

  3. anonymous
    • 5 years ago
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    Really though tell them to ease up a little. They are beating up my brain......

  4. nowhereman
    • 5 years ago
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    Any concrete question?

  5. anonymous
    • 5 years ago
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    yeah. I am having a hard time understanding this concept of going down the list of axioms to prove whether or not they make the statement true.

  6. anonymous
    • 5 years ago
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    here is an example: determine whether the given set of vectors is closed under addition and closed under scalar multiplication. The set of scalars is the set of all real numbers. The set of all solutions to the diff eqn. y'+9y=ax^2

  7. nowhereman
    • 5 years ago
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    Well, there are a lot of nice properties of and theorems relating to vector spaces. To know that they apply you have to know whether some set really is a vector space. And for that all those axioms have to be true. Basically it's a kind of linear structure.

  8. anonymous
    • 5 years ago
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    Yeah, thats sounds similar to what my teacher says that you have to show that they are all true. I am having trouble proving the axioms. I guess I just have to look at examples and figure it out.

  9. nowhereman
    • 5 years ago
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    For solutions of differential equations you normally need a homogeneous form. In your example that is not the case and you can see that the solutions don't form a vector space.

  10. anonymous
    • 5 years ago
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    For the problem, the ax^2 is really 4x^2. sorry for the typo.

  11. anonymous
    • 5 years ago
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    oh ok.

  12. nowhereman
    • 5 years ago
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    Still the same. Just asume you have a solution f: x ↦ y and look whether for any c the solution c*f is still a solution. You will see, it's not.

  13. anonymous
    • 5 years ago
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    so if it were to equal zero than it would be one?

  14. nowhereman
    • 5 years ago
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    yes, because the differentiation is linear as you from the basic differentiation rules.

  15. anonymous
    • 5 years ago
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    O.k thanks. Your a life saver.

  16. nowhereman
    • 5 years ago
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    You're welcome :-)

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